why was it difficult for the woman to cross the road​

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Explanation:

The phrasing "2/3 of 4/5 of his 2nd exam on his 3rd test" is a bit clunky in my opinion. It seems more complicated than it has to be.

The student got 80 on the second exam. 4/5 of this is (4/5)*80 = 0.8*80 = 64. Then we take 2/3 of this to get (2/3)*64 = 42.667 approximately. If we assume only whole number scores are given, then this would round to 43.

Let x be the score on the fourth exam. Since 5 points of extra credit are given, the student actually got x+5 points on this exam.

So we have these scores

Adding up these scores and dividing by 4 will get us the average

(sum of scores)/(number of scores) = average

(70+80+43+x+5)/4 = 80

(x+198)/4 = 80

x+198 = 4*80

x+198 = 320

x = 320 - 198

x = 122

So the student got a score of x+5 = 122+5 = 127 on the fourth exam.

Answer:

2w<23

Step-by-step explanation:

The product of w and 2 mean that w multiplied by 2

Answer:

C. 3rd quartile

Step-by-step explanation:

Percentile:

A data belonging to the xth percentile means that the data is greater than x% of the values of the data-set, and smaller than (100 - x)%.

Point below 75% of the cases lie:

This is the 75th percentile, which is the 3rd quartile, as 75 = 3*100/4. Thus, the correct answer is given by option c.

Step-by-step explanation:

k=8.6

i think it is the right answer

Answer:

23

Step-by-step explanation:

Answer:

0.1143 = 11.43% probability that all but one of them are using Chrome as their browser

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they use Chrome, or they do not. The probability of a person using Chrome is independent of any other person, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

Chrome: 52.57%

This means that [tex]p = 0.5257[/tex]

Sample of 6 people

This means that [tex]n = 6[/tex]

What is the probability that all but one of them are using Chrome as their browser?

5 using Chrome, so:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 5) = C_{6,5}.(0.5257)^{5}.(1-0.5257)^{1} = 0.1143[/tex]

0.1143 = 11.43% probability that all but one of them are using Chrome as their browser

Answer:

The gradient is -0.25

Step-by-step explanation:

Given

[tex](x_1,y_1) = (-5,3)[/tex]

[tex](x_2,y_2) = (5,0.5)[/tex]

Required

The gradient (m)

This is calculated as:

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

So, we have:

[tex]m = \frac{0.5-3}{5--5}[/tex]

[tex]m = \frac{-2.5}{10}[/tex]

[tex]m = -0.25[/tex]

Answer:

if the sum of two angles is 90° they are said to be complementary

If one angle is x

the other should be (90 - x)°

so the complementary angle of x is (90 - x)

Answer:

Andres

why?

Because he is median salary for cpa

The answer for the first line segment : (-3,-7) (-4,0)

The answer for 2nd line segment is :(-3,8) (-9,-5)

Step-by-step explanation:

Let do line segment QR and ST. first.

Step 1: Find a line that contains a points that is perpendicular to the line of reflection

"A reflection of a pre image and new image is perpendicular to the line of reflection.

This means for points Q,S,T and R, there is a line that. contains one point that is perpendicular to the line of reflection.

A line that is perpendicular to the line of reflection is the negative reciprocal of the slope so this means all 4 lines must be on a different slopes but the slopes must be 1/2.

To simplify, things, here are the lines that will all 4 points be on

Step 2: Find a point where both the line and line of reflection intersect at.

Now we need to find a line where both the line of reflections and the 4 lines will intersect at separately.

Step 3: Find the endpoints given the midpoint and the originally endpoint.

A reflection per and new image is equidistant from the point of reflection. So we. an say that the point where the line intersect is the midpoint of the pre and new image.

Using this info,

Connect R prime and Q prime. And that the new line segments

Connects S prime and T prime and that the new line segments.

[tex]\huge\textsf{Hey there!}[/tex]

[tex]\mathsf{3^3 \times 5^4}[/tex]

[tex]\mathsf{3^3}[/tex]

[tex]\mathsf{= 3\times3\times3}[/tex]

[tex]\mathsf{= 9\times3}[/tex]

[tex]\mathsf{= \bf 27}[/tex]

[tex]\mathsf{5^4}[/tex]

[tex]\mathsf{= 5\times 5\times5\times 5}[/tex]

[tex]\mathsf{= 25\times25}[/tex]

[tex]\mathsf{= \bf 625}[/tex]

[tex]\mathsf{27 \times625}[/tex]

[tex]\mathsf{= \bf 16,875}[/tex]

[tex]\mathsf{2\times5^3\times11}[/tex]

[tex]\mathsf{5^3}[/tex]

[tex]\mathsf{= 5\times 5\times5}[/tex]

[tex]\mathsf{= 25\times 5}[/tex]

[tex]\mathsf{\bf = 125}[/tex]

[tex]\mathsf{2\times125\times11}[/tex]

[tex]\mathsf{= 250\times11}[/tex]

[tex]\mathsf{\bf = 2,750}[/tex]

[tex]\large\textsf{Find the Greatest Common Factor (GCF) of 16,875 \& 2,750}[/tex]

[tex]\large\textsf{16,875: 1, 3, 5, 9, 15, 25, 27, 45, 75, 125, 135, 225, 375, 625, 675, 1,125,}\\\\\large\textsf{1,875, 3,375, 5,625, \& 16,875}[/tex]

[tex]\large\textsf{2,750: 1, 2, 5, 10, 11, 22, 25, 50, 55, 110, 125, 250, 275, 550, 1,375, \& 2,750}[/tex]

[tex]\boxed{\boxed{\huge\textsf{Answer: the GCF is \bf 125 }}}\huge\checkmark[/tex]

[tex]\large\textsf{Good luck on your assignment and enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Answer:

B. Less than 1

Step-by-step explanation:

You could plug in values of n greater than 1 and see what happens....

Example n=2 gives |(.5+.2i)^2|

Simplifying inside gives |(.5)^2+2(.5)(.2i)+(.2i)^2|

=|.25+.2i+.04i^2|=|.25+.2i-.04|=|.21+.2i|.

Applying the absolute value part gives sqrt(.21^2+.2^2)=sqrt(.0441+.04)=sqrt(.0841)=.29

This value is less than 1.

We should also be able to do the absolute value first then the power.

|.5+.2i|=sqrt(.25+.04)=sqrt(.29)

So |.5+.2i|^2=.29 which is what we got long way around.

Anyways (sqrt(.29))^n where n is greater than 1 will result in a number greater than 0 but less than 1.

Answer:

compound interest

Step-by-step explanation:

compound interest yields higher profit than simple interest

Answer:

the simple interest function will be greater in the beginning, but the

compound interest equation will overtake the simple after a while.

it appears that the 4% compound equation will overtake the simple at about 10 years

Step-by-step explanation:

[tex]1 + .05t = 1(1.04)^t[/tex]

Answer in fraction form is 1/3

Answer in decimal form is 0.3333

Pick one answer only.

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Explanation:

The sample space is the set of all possible outcomes. In this case, the outcomes consist of values between 1 and 6

S = sample space

S = {1,2,3,4,5,6}

There are 6 items here. Let B = 6.

We want to roll a number greater than 4, so the event space we're after is

E = {5,6}

which consists of 2 items. Let A = 2.

The probability we want is A/B = 2/6 = 1/3 = 0.3333

So if you go with the fraction option, then you'll type in 1/3

If you go with the decimal option, then you'll type in 0.3333

Answer:

Dotted Curve

Shaded Area includes the Origin.

Step-by-step explanation:

I am assuming that you mean 4x² > y-3.

Since the inequality sign is greater than, the line would be dotted. The curve would be solid if the inequality was an 'X or equal to' sign.

----------------------------------

4(0)² > 0 - 3

0 > -3

Since the given statement is true, the origin is a solution and we would shade below the curve.

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See the graph attached.

Hope this helps.

Answer:

a)  b^5/12

Step-by-step explanation:

b^2/3 ÷ b^1/4    [if bases are the same then subtract the exponents]

2/3 - 1/4 = 5/12

Step-by-step explanation:

BUC={2,3,4,5,6,7}

An(BUC)={4,6,7}

AnB={4,7},AnC={6,7}

(AnB)U(AnC)={4,6,7}

Au(BnC)={1,4,5,6,7}

(AUB)n(AUC)= {1,4,5,6,7}

Answer: hello there here is your answer:

Still air speed:450 mph.

Step-by-step explanation:

500-450=450-400=50 mph

Still air speed:450 mph. Wind speed:50 mph..

hope this help have a good day

Answer:

[tex]\bar x = 107.11[/tex]

[tex]\sigma_x = 31.07[/tex]

Step-by-step explanation:

See comment for complete question

Given

[tex]x: 97\ 178\ 129\ 90\ 75\ 94\ 116\ 100\ 85[/tex]

Solving (a): The sample mean

This is calculated using:

[tex]\bar x = \frac{\sum x}{n}[/tex]

So, we have:

[tex]\bar x = \frac{97+ 178+ 129+ 90+ 75+ 94+ 116+ 100+ 85}{9}[/tex]

[tex]\bar x = \frac{964}{9}[/tex]

[tex]\bar x = 107.11[/tex]

Solving (b): The sample standard deviation

This is calculated as:

[tex]\sigma_x = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}[/tex]

So, we have:

[tex]\sigma_x = \sqrt{\frac{(97 - 107.11)^2 +.............+ (85- 107.11)^2 }{9-1}}[/tex]

[tex]\sigma_x = \sqrt{\frac{7720.8889}{8}}[/tex]

[tex]\sigma_x = \sqrt{965.1111125}[/tex]

[tex]\sigma_x = 31.07[/tex]

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Answer:

a) P(man asked for a raise) = 0.21.

b) P(man received raise, given he asked for one) = 0.6.

c) P(man asked for raise and received raise) = 0.126.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

Question a:

21% asked for a raise, so:

P(man asked for a raise) = 0.21.

Question b:

Event A: Asked for a raise.

Event B: Received a raise:

21% had asked for a raise and 60% of the men who had asked for a raise received the raise:

This means that [tex]P(A) = 0.21, P(A \cap B) = 0.21*0.6[/tex], thus:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.21*0.6}{0.6} = 0.6[/tex]

P(man received raise, given he asked for one) = 0.6.

Question c:

[tex]P(A \cap B) = 0.21*0.6 = 0.126[/tex]

P(man asked for raise and received raise) = 0.126.

Answer:

numerator degree of freedom = 3

Denominator degree of freedom = 47

Step-by-step explanation:

The numerator degree of freedom is given by :

p - 1 ; where p = number of predictors ;

p = number of independent variables + 1

Number of independent variables = 3

p = 3 + 1 = 4

Numerator degree of freedom = p - 1 = 4 - 1 = 3

The denominator degree of freedom = n - p ; where n = number of observations

Number of observations, n = 51

Denominator degree of freedom = n - p = 51 - 4 = 47

Answer:

0.7123 = 71.23% probability that a random sample of 75 detached houses in Franklin County had a mean price greater than $190,000 in 2009.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean price for a detached house in Franklin County, OH in 2009 was $192,723. Suppose we know that the standard deviation was $42,000.

This means that [tex]\mu = 192723, \sigma = 42000[/tex]

Sample of 75:

This means that [tex]n = 75, s = \frac{42000}{\sqrt{75}}[/tex]

What is the probability that a random sample of 75 detached houses in Franklin County had a mean price greater than $190,000 in 2009?

1 subtracted by the p-value of Z when X = 190000. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{190000 - 192723}{\frac{42000}{\sqrt{75}}}[/tex]

[tex]Z = -0.56[/tex]

[tex]Z = -0.56[/tex] has a p-value of 0.2877

1 - 0.2877 = 0.7123

0.7123 = 71.23% probability that a random sample of 75 detached houses in Franklin County had a mean price greater than $190,000 in 2009.

Answer: 27 Dresses and 50 Pants

Step-by-step explanation:

If she sold 9 pairs of pants and

9 x 2 = 18

18 + 9 = 27

20 + 10 = 30

30 + 20 = 50

Evelyn sold 9 dresses and 20 pairs of pants on Friday, and on Saturday, she sold 18 dresses and 30 pairs of pants.

Evelyn's sales of dresses and pants over two days, Friday and Saturday. We'll use some mathematical expressions and reasoning to find out how many dresses Evelyn sold on each day.

Let's start by assigning some variables to represent the number of dresses and pants Evelyn sold on Friday and Saturday. We'll use "F" for Friday and "S" for Saturday. So, let [tex]D_F[/tex] be the number of dresses sold on Friday, [tex]D_S[/tex] be the number of dresses sold on Saturday, [tex]P_F[/tex] be the number of pants sold on Friday, and [tex]P_S[/tex] be the number of pants sold on Saturday.

According to the problem, on Friday, Evelyn sold 9 dresses, which can be expressed as:

[tex]D_F[/tex] = 9

She also sold 20 pairs of pants on Friday:

[tex]P_F[/tex] = 20

Now, let's move on to Saturday's sales. It says she sold twice as many dresses as Friday, which means the number of dresses on Saturday is double that of Friday's sales:

[tex]D_S = 2 * D_F[/tex]

Additionally, she sold 10 more pairs of pants on Saturday compared to Friday:

[tex]P_S = P_F + 10[/tex]

We already know that [tex]D_F = 9[/tex], so we can find the number of dresses sold on Saturday by substituting this value into the equation for [tex]D_S[/tex]:

[tex]D_S = 2 * 9 = 18[/tex]

Next, we'll calculate the number of pants sold on Saturday using the given information. Since [tex]P_F = 20[/tex], we can find [tex]P_S[/tex]:

[tex]P_S = 20 + 10 = 30[/tex]

So, to summarize, Evelyn sold 9 dresses and 20 pairs of pants on Friday, and on Saturday, she sold 18 dresses and 30 pairs of pants.

To know more about Equation here

https://brainly.com/question/15977368

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Answer:

assuming that you start at the origin (0,0)

(-4,-1) would be the poiny

x coord = -4

y coord = -1

the point is in the 3 quadrant

Step-by-step explanation:

12.8, pythagorean theorem.

Answer:

A a vertex

Step-by-step explanation:

When rays meet they form a point is formed known as the vertex.

Answer:

Kindly check explanation

Step-by-step explanation:

A.)

H0 : μ = 5000

H0 : μ > 5000

xbar = 6671 ; s = 8185.21 ; n = 52 ; α = 0.05

The test statistic :

T = (xbar - μ) ÷ (s/√(n))

T = (6671 - 5000) ÷ (8185.21/√52)

T = 3.814

Pvalue from Test statistic : ; df = n - 1 = 52-1= 51

Pvalue at 3.814; 51 = 0.000185

Pvalue < α ; Reject H0 and conclude that average birth is greater than 5000

B)

H0 : μ = 6000

H0 : μ < 6000

xbar = 4187 ; s = 4386 ; n = 52 ; α = 0.01

The test statistic :

T = (xbar - μ) ÷ (s/√(n))

T = (4187 - 6000) ÷ (4386/√52)

T = - 2.981

Pvalue from Test statistic : ; df = n - 1 = 52-1= 51

Pvalue at - 2.981; 51 = 0.0022

Pvalue < α ; Reject H0 and conclude that average death is less than 6000

C.)

H0 : μ < 2500

H0 : μ ≥ 2500

xbar = 2744 ; s = 3134.41 ; n = 52 ; α = 0.05

The test statistic :

T = (xbar - μ) ÷ (s/√(n))

T = (2744 - 2500) ÷ (3134.41/√52)

T = 0.561

Pvalue from Test statistic : ; df = n - 1 = 52-1= 51

Pvalue at 0.561; 51 = 0.289

Pvalue > α ; Fail to Reject H0 and conclude that average marriage is not greater Tha or equal to 2500

D.)

H0 : μ = 4000

H0 : μ ≤ 4000

xbar = 1451 ; s = 1217 ; n = 52 ; α = 0.01

The test statistic :

T = (xbar - μ) ÷ (s/√(n))

T = (1451 - 4000) ÷ (1217/√52)

T = - 15.10

Pvalue from Test statistic : ; df = n - 1 = 52-1= 51

Pvalue at - 15.10; 51 = 0.000001

Pvalue < α ; Reject H0 and conclude that average divorce is less eqaul to 4000